A. Rate of Change · Derivatives and Differentiation Rules

Combining the Rules: Radical and Rational Functions

A3.5About 75 minutes

Learning goals

I can choose which differentiation rule to apply first by identifying the last operation in a function.
I can combine the product, quotient and chain rules to differentiate polynomial, rational and radical functions.
I can factor a derivative to find horizontal tangents.
I can find second derivatives of rational and radical functions.

Words to know

outermost operation
The last operation you would perform to evaluate a function; it decides which rule to apply first.

common factor
A factor that appears in every term, such as (3x − 1)³ in 2x(3x − 1)⁴ + 12x²(3x − 1)³.

domain
The set of x-values for which a function is defined; radicals and denominators restrict it.

vertical tangent
A tangent line that is vertical; the derivative is undefined there because the slope grows without bound.

All the tools at once

You now have a full toolkit: the power rule (for any rational exponent), the constant multiple, sum and difference rules, the product rule, the quotient rule and the chain rule. Real functions often need several of them at once, such as y = x²(3x − 1)⁴ or y = x/√(x² + 4).

The skill in this lesson is choosing the order. Work from the outside in: decide which rule applies to the whole function, apply it, and handle each inside piece with the rules it needs.

A strategy for any function: look at the last operation you would do if you evaluated it with a calculator. That operation tells you which rule to use first. Then repeat the question for each piece inside.

How much to simplify

The curriculum's focus is on applying the rules correctly, not on heavy simplification. Simplify enough to answer the question: factor out common factors if you need to solve y′ = 0, but if you only need a slope at one point, substitute right away.

Worked example

Example 1: Product and chain

Differentiate y = x²(3x − 1)⁴ and find where the tangent is horizontal.

Last operation: multiplication. Product rule with f = x² and g = (3x − 1)⁴.
g′ needs the chain rule: g′ = 4(3x − 1)³ · 3 = 12(3x − 1)³.
y′ = 2x(3x − 1)⁴ + x² · 12(3x − 1)³.
Factor out the common factor 2x(3x − 1)³: y′ = 2x(3x − 1)³[(3x − 1) + 6x] = 2x(3x − 1)³(9x − 1).
y′ = 0 when x = 0, x = 1/3 or x = 1/9.

y′ = 2x(3x − 1)³(9x − 1); horizontal tangents at x = 0, 1/9 and 1/3. Factoring made the zeros easy to read.

Worked example

Example 2: A product with a radical

Differentiate y = x√(1 − x²) and find its horizontal tangents.

Product rule with f = x and g = (1 − x²)^(1/2).
Chain rule: g′ = (1/2)(1 − x²)^(−1/2)(−2x) = −x/√(1 − x²).
y′ = 1 · √(1 − x²) + x · (−x/√(1 − x²)).
Common denominator √(1 − x²): y′ = [(1 − x²) − x²]/√(1 − x²) = (1 − 2x²)/√(1 − x²).
y′ = 0 when 1 − 2x² = 0, x = ±1/√2. Then y = (1/√2)√(1/2) = 1/2 or y = −1/2.

y′ = (1 − 2x²)/√(1 − x²); horizontal tangents at (1/√2, 1/2) and (−1/√2, −1/2).

y = x√(1 − x²) on its domain −1 ≤ x ≤ 1. Its derivative is 0 where 1 − 2x² = 0 (horizontal tangents at x = ±1/√2) and undefined at x = ±1, where the tangents are vertical.

Worked example

Example 3: Chain and quotient

Differentiate y = [(x + 1)/(x − 1)]³.

Last operation: cubing. Chain rule: y′ = 3[(x + 1)/(x − 1)]² · d/dx [(x + 1)/(x − 1)].
Quotient rule for the inside: [(x − 1) − (x + 1)]/(x − 1)² = −2/(x − 1)².
y′ = 3(x + 1)²/(x − 1)² · (−2)/(x − 1)² = −6(x + 1)²/(x − 1)⁴.

y′ = −6(x + 1)²/(x − 1)⁴

Worked example

Example 4: A rational function with a radical

Differentiate y = x/√(x² + 4).

Rewrite as a product: y = x(x² + 4)^(−1/2).
Product and chain rules: y′ = (x² + 4)^(−1/2) + x · (−1/2)(x² + 4)^(−3/2)(2x) = (x² + 4)^(−1/2) − x²(x² + 4)^(−3/2).
Factor out (x² + 4)^(−3/2): y′ = (x² + 4)^(−3/2)[(x² + 4) − x²] = 4(x² + 4)^(−3/2).
So y′ = 4/(x² + 4)^(3/2), which is always positive.

y′ = 4/(x² + 4)^(3/2): the function is increasing everywhere. (The quotient rule gives the same result with a little more algebra.)

Worked example

Example 5: A second derivative

Find y″ for y = 1/(x + 1).

y = (x + 1)⁻¹, so y′ = −(x + 1)⁻².
Differentiate again (chain rule, inner derivative 1): y″ = 2(x + 1)⁻³.

y″ = 2/(x + 1)³ (for x ≠ −1). You will use second derivatives like this to describe the shape of rational graphs in Curve Sketching: Connecting f, f′ and f″.

Just for fun

A calculus student was asked to differentiate a function with a product, a quotient and three chain rules nested inside one another. Their answer began, "Once upon a time, there was an outer function…"

Hands-on activity

Rule detective cards

You need: 12 index cards or scraps of paper, a pencil and a timer (a phone works).

Write one function on each card. Use these: x³ − 4x, x²(x + 1)³, (2x + 1)/(x − 4), √(x² + 1), x√(x + 3), 1/(x² + 5), (x² + 1)⁵(x − 2), [(x − 1)/(x + 1)]², 5x^(2/3), (x + 1)/√x, ∛(1 − x³), x/(x + 1)².
Shuffle. For each card, say aloud the first rule you would use (term by term, product, quotient, chain) and then the rules needed inside. Time yourself.
Pick four cards and fully differentiate them. Check each answer by the centred estimate at x = 1 (or x = 2 if x = 1 is not in the domain).

What to look for: first rules: term by term; product; quotient; chain; product; chain (or quotient); product; chain; power rule; quotient (or divide first: x^(1/2) + x^(−1/2)); chain; quotient (or product with (x + 1)⁻²). Differences of opinion are fine where two routes both work.

Your turn

Differentiate y = x³(2x + 5)².

Differentiate y = (x − 1)²/(x + 2).

Differentiate y = √(x/(x + 1)) and find the slope at x = 1.

Differentiate y = (x² + 1)³(x − 2)⁴. (Leave factored.)

Find the slope of y = x√(x + 3) at x = 1.

Find y″ for y = (2x + 1)⁵.

Which rule do you apply first to y = (x² + 1)/(x − 3)²? Differentiate it.

More practice: warm-up

First rule for y = (x² + 3)⁷?

First rule for y = x²√x?

First rule for y = (x + 1)(x − 1)⁵?

Differentiate y = (4 − x)³.

Differentiate y = 2/(3x + 1).

More practice: on your own

Differentiate y = (x + 2)³(x − 1)² and find the x-values of the horizontal tangents.

Differentiate y = x²/√(x² + 1).

Find the equation of the tangent to y = (x + 1)√(2x − 1) at x = 1.

Find y″ for y = √(x + 1) and evaluate at x = 3.

Show that y = (x − 3)/√(x² + 9) has a horizontal tangent at x = −3.

Challenge

Differentiate y = √(x + √x).

A circular oil spill on a lake (a cleanup drill) has radius r(t) = 20√(t + 1) metres after t hours. How fast is its area growing at t = 3?

Find all x where the tangent to y = x(x − 4)³ is horizontal, and classify each point using the sign of y′.

Check your understanding

Quiz

Which rule do you apply first to y = (x³ − 1)⁴(2x + 7)?

What is the derivative of y = (5x − 2)⁻¹?

For y = x(x + 1)³, which expression is y′?

What is y′ for y = √(x² + 16) at x = 3?

y′ = 2x(3x − 1)³(9x − 1). How many horizontal tangents does y have?

If you only need the slope at x = 2, the best plan is to:

For parents and teachers

Big idea (A3.5): most functions need several rules together. The key skill is choosing which rule to use first: look at the last operation you would do to evaluate the function (sum, product, quotient or power of a function), apply that rule, then work inward.

How to help: Point at a function and ask "What's the last thing you'd do on a calculator?" before your teen differentiates. Encourage them to write down f, f′, g and g′ separately before combining them.

Common difficulties:

Forgetting an inner chain-rule factor inside a product or quotient.
Algebra errors when factoring out common powers such as (3x − 1)³ or (x² + 4)^(−3/2): the smaller exponent comes out.
Over-simplifying when only a numerical slope is needed.
Missing the domain restrictions of radicals (for example, x√(1 − x²) needs −1 ≤ x ≤ 1).